(a) If (log_{10} 2 = 0.3010) and (log_{10} 3 = 0.4771), calculate without using tables, the value of (log_{10} 0.72).
(b) A hawk on top of a tree, 20 metres high views a chick on the ground at an angle of depression of 39°. Find, correct to 2 significant figures, the distance of the chick from the bottom of the tree.
Explanation
(a) (log_{10} 0.72 =log_{10} (frac{72}{100}))
= (log_{10} 72 – log_{10} 100)
= (72 = 2^{3} times 3^{2})
(log_{10} 72 = log_{10} 2^{3} + log_{10} 3^{2} = 3log_{10} 2 + 2 log_{10} 3)
= (3(0.3010) + 2(0.4771) = 0.9030 + 0.9542 = 1.8572)
(therefore log_{10} 0.72 equiv log_{10} 72 – log_{10} 100)
= (1.8572 – 2 = -0.1428)
(b) (tan 39° = frac{20}{AB} )
= (frac{20}{tan 39°})
(AB = 24.698 m approxeq 25m)